Left Termination proof of /tmp/tmpow

Left Termination of the query pattern sort(b,f) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

sort₂(X, Y) :- perm₂(X, Y), sorted₁(Y).
sorted₁({}₀).
sorted₁(.₂(X, {}₀)).
sorted₁(.₂(X, .₂(Y, Z))) :- le₂(X, Y), sorted₁(.₂(Y, Z)).
perm₂({}₀, {}₀).
perm₂(.₂(X, .₂(Y, {}₀)), .₂(U, .₂(V, {}₀))) :- delete₃(U, .₂(X, Y), Z), perm₂(Z, V).
delete₃(X, .₂(X, Y), Y).
delete₃(X, .₂(Y, Z), W) :- delete₃(X, Z, W).
le₂(s₁(X), s₁(Y)) :- le₂(X, Y).
le₂(0₀, s₁(X)).
le₂(0₀, 0₀).

With regard to the inferred argument filtering the predicates were used in the following modes:
sort₂: (b,f)
perm₂: (b,f)
delete₃: (f,b,f)
sorted₁: (b)
le₂: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

SORT_2_IN_GA₂(X, Y) -> IF_SORT_2_IN_1_GA₃(X, Y, perm_2_in_ga₂(X, Y))
SORT_2_IN_GA₂(X, Y) -> PERM_2_IN_GA₂(X, Y)
PERM_2_IN_GA₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
PERM_2_IN_GA₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> DELETE_3_IN_AGA₃(U, ._2₂(X, Y), Z)
DELETE_3_IN_AGA₃(X, ._2₂(Y, Z), W) -> IF_DELETE_3_IN_1_AGA₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
DELETE_3_IN_AGA₃(X, ._2₂(Y, Z), W) -> DELETE_3_IN_AGA₃(X, Z, W)
IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> IF_PERM_2_IN_2_GA₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> PERM_2_IN_GA₂(Z, V)
IF_SORT_2_IN_1_GA₃(X, Y, perm_2_out_ga₂(X, Y)) -> IF_SORT_2_IN_2_GA₃(X, Y, sorted_1_in_g₁(Y))
IF_SORT_2_IN_1_GA₃(X, Y, perm_2_out_ga₂(X, Y)) -> SORTED_1_IN_G₁(Y)
SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_in_gg₂(X, Y))
SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> LE_2_IN_GG₂(X, Y)
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LE_2_IN_1_GG₃(X, Y, le_2_in_gg₂(X, Y))
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> IF_SORTED_1_IN_2_G₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> SORTED_1_IN_G₁(._2₂(Y, Z))

The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

The argument filtering Pi contains the following mapping:
sort_2_in_ga₂(x1, x2) = sort_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
if_sort_2_in_1_ga₃(x1, x2, x3) = if_sort_2_in_1_ga₁(x3)
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_1_ga₁(x5)
delete_3_in_aga₃(x1, x2, x3) = delete_3_in_aga₁(x2)
delete_3_out_aga₃(x1, x2, x3) = delete_3_out_aga₂(x1, x3)
if_delete_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_delete_3_in_1_aga₁(x5)
if_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x3, x6)
if_sort_2_in_2_ga₃(x1, x2, x3) = if_sort_2_in_2_ga₂(x2, x3)
sorted_1_in_g₁(x1) = sorted_1_in_g₁(x1)
sorted_1_out_g₁(x1) = sorted_1_out_g
if_sorted_1_in_1_g₄(x1, x2, x3, x4) = if_sorted_1_in_1_g₃(x2, x3, x4)
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
if_sorted_1_in_2_g₄(x1, x2, x3, x4) = if_sorted_1_in_2_g₁(x4)
sort_2_out_ga₂(x1, x2) = sort_2_out_ga₁(x2)
IF_SORTED_1_IN_2_G₄(x1, x2, x3, x4) = IF_SORTED_1_IN_2_G₁(x4)
SORT_2_IN_GA₂(x1, x2) = SORT_2_IN_GA₁(x1)
IF_DELETE_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_DELETE_3_IN_1_AGA₁(x5)
IF_SORT_2_IN_1_GA₃(x1, x2, x3) = IF_SORT_2_IN_1_GA₁(x3)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_1_GA₁(x5)
SORTED_1_IN_G₁(x1) = SORTED_1_IN_G₁(x1)
IF_SORT_2_IN_2_GA₃(x1, x2, x3) = IF_SORT_2_IN_2_GA₂(x2, x3)
IF_LE_2_IN_1_GG₃(x1, x2, x3) = IF_LE_2_IN_1_GG₁(x3)
DELETE_3_IN_AGA₃(x1, x2, x3) = DELETE_3_IN_AGA₁(x2)
LE_2_IN_GG₂(x1, x2) = LE_2_IN_GG₂(x1, x2)
IF_SORTED_1_IN_1_G₄(x1, x2, x3, x4) = IF_SORTED_1_IN_1_G₃(x2, x3, x4)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_GA₂(x3, x6)

We have to consider all (P,R,Pi)-chains

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

SORT_2_IN_GA₂(X, Y) -> IF_SORT_2_IN_1_GA₃(X, Y, perm_2_in_ga₂(X, Y))
SORT_2_IN_GA₂(X, Y) -> PERM_2_IN_GA₂(X, Y)
PERM_2_IN_GA₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
PERM_2_IN_GA₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> DELETE_3_IN_AGA₃(U, ._2₂(X, Y), Z)
DELETE_3_IN_AGA₃(X, ._2₂(Y, Z), W) -> IF_DELETE_3_IN_1_AGA₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
DELETE_3_IN_AGA₃(X, ._2₂(Y, Z), W) -> DELETE_3_IN_AGA₃(X, Z, W)
IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> IF_PERM_2_IN_2_GA₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> PERM_2_IN_GA₂(Z, V)
IF_SORT_2_IN_1_GA₃(X, Y, perm_2_out_ga₂(X, Y)) -> IF_SORT_2_IN_2_GA₃(X, Y, sorted_1_in_g₁(Y))
IF_SORT_2_IN_1_GA₃(X, Y, perm_2_out_ga₂(X, Y)) -> SORTED_1_IN_G₁(Y)
SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_in_gg₂(X, Y))
SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> LE_2_IN_GG₂(X, Y)
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LE_2_IN_1_GG₃(X, Y, le_2_in_gg₂(X, Y))
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> IF_SORTED_1_IN_2_G₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> SORTED_1_IN_G₁(._2₂(Y, Z))

The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LE_2_IN_GG₂}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> SORTED_1_IN_G₁(._2₂(Y, Z))
SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> SORTED_1_IN_G₁(._2₂(Y, Z))
SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> IF_SORTED_1_IN_1_G₄(X, Y, Z, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
SORTED_1_IN_G₁(x1) = SORTED_1_IN_G₁(x1)
IF_SORTED_1_IN_1_G₄(x1, x2, x3, x4) = IF_SORTED_1_IN_1_G₃(x2, x3, x4)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

IF_SORTED_1_IN_1_G₃(Y, Z, le_2_out_gg) -> SORTED_1_IN_G₁(._2₂(Y, Z))
SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> IF_SORTED_1_IN_1_G₃(Y, Z, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₁(le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg
if_le_2_in_1_gg₁(le_2_out_gg) -> le_2_out_gg

The set Q consists of the following terms:

le_2_in_gg₂(x0, x1)
if_le_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SORTED_1_IN_G₁, IF_SORTED_1_IN_1_G₃}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

SORTED_1_IN_G₁(._2₂(X, ._2₂(Y, Z))) -> IF_SORTED_1_IN_1_G₃(Y, Z, le_2_in_gg₂(X, Y))

The remaining Dependency Pairs were at least non-strictly be oriented.

IF_SORTED_1_IN_1_G₃(Y, Z, le_2_out_gg) -> SORTED_1_IN_G₁(._2₂(Y, Z))

With the implicit AFS there is no usable rule.

Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 0
POL(SORTED_1_IN_G₁(x₁)) = x₁
POL(._2₂(x₁, x₂)) = 1 + x₂
POL(le_2_out_gg) = 0
POL(IF_SORTED_1_IN_1_G₃(x₁, x₂, x₃)) = 1 + x₂
POL(le_2_in_gg₂(x₁, x₂)) = 0
POL(s_1₁(x₁)) = 0
POL(if_le_2_in_1_gg₁(x₁)) = 0

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

IF_SORTED_1_IN_1_G₃(Y, Z, le_2_out_gg) -> SORTED_1_IN_G₁(._2₂(Y, Z))

The TRS R consists of the following rules:

le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₁(le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg
if_le_2_in_1_gg₁(le_2_out_gg) -> le_2_out_gg

The set Q consists of the following terms:

le_2_in_gg₂(x0, x1)
if_le_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SORTED_1_IN_G₁, IF_SORTED_1_IN_1_G₃}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

DELETE_3_IN_AGA₃(X, ._2₂(Y, Z), W) -> DELETE_3_IN_AGA₃(X, Z, W)

The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

DELETE_3_IN_AGA₃(X, ._2₂(Y, Z), W) -> DELETE_3_IN_AGA₃(X, Z, W)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
DELETE_3_IN_AGA₃(x1, x2, x3) = DELETE_3_IN_AGA₁(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

DELETE_3_IN_AGA₁(._2₂(Y, Z)) -> DELETE_3_IN_AGA₁(Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {DELETE_3_IN_AGA₁}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

DELETE_3_IN_AGA₁(._2₂(Y, Z)) -> DELETE_3_IN_AGA₁(Z)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> PERM_2_IN_GA₂(Z, V)
PERM_2_IN_GA₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))

The TRS R consists of the following rules:

sort_2_in_ga₂(X, Y) -> if_sort_2_in_1_ga₃(X, Y, perm_2_in_ga₂(X, Y))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))
delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)
if_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_in_ga₂(Z, V))
if_perm_2_in_2_ga₆(X, Y, U, V, Z, perm_2_out_ga₂(Z, V)) -> perm_2_out_ga₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0)))
if_sort_2_in_1_ga₃(X, Y, perm_2_out_ga₂(X, Y)) -> if_sort_2_in_2_ga₃(X, Y, sorted_1_in_g₁(Y))
sorted_1_in_g₁([]_0) -> sorted_1_out_g₁([]_0)
sorted_1_in_g₁(._2₂(X, []_0)) -> sorted_1_out_g₁(._2₂(X, []_0))
sorted_1_in_g₁(._2₂(X, ._2₂(Y, Z))) -> if_sorted_1_in_1_g₄(X, Y, Z, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(X)) -> le_2_out_gg₂(0_0, s_1₁(X))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_sorted_1_in_1_g₄(X, Y, Z, le_2_out_gg₂(X, Y)) -> if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_in_g₁(._2₂(Y, Z)))
if_sorted_1_in_2_g₄(X, Y, Z, sorted_1_out_g₁(._2₂(Y, Z))) -> sorted_1_out_g₁(._2₂(X, ._2₂(Y, Z)))
if_sort_2_in_2_ga₃(X, Y, sorted_1_out_g₁(Y)) -> sort_2_out_ga₂(X, Y)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), Z)) -> PERM_2_IN_GA₂(Z, V)
PERM_2_IN_GA₂(._2₂(X, ._2₂(Y, []_0)), ._2₂(U, ._2₂(V, []_0))) -> IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), Z))

The TRS R consists of the following rules:

delete_3_in_aga₃(X, ._2₂(X, Y), Y) -> delete_3_out_aga₃(X, ._2₂(X, Y), Y)
delete_3_in_aga₃(X, ._2₂(Y, Z), W) -> if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_in_aga₃(X, Z, W))
if_delete_3_in_1_aga₅(X, Y, Z, W, delete_3_out_aga₃(X, Z, W)) -> delete_3_out_aga₃(X, ._2₂(Y, Z), W)

The argument filtering Pi contains the following mapping:
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
delete_3_in_aga₃(x1, x2, x3) = delete_3_in_aga₁(x2)
delete_3_out_aga₃(x1, x2, x3) = delete_3_out_aga₂(x1, x3)
if_delete_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_delete_3_in_1_aga₁(x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_1_GA₁(x5)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

Q DP problem:
The TRS P consists of the following rules:

IF_PERM_2_IN_1_GA₁(delete_3_out_aga₂(U, Z)) -> PERM_2_IN_GA₁(Z)
PERM_2_IN_GA₁(._2₂(X, ._2₂(Y, []_0))) -> IF_PERM_2_IN_1_GA₁(delete_3_in_aga₁(._2₂(X, Y)))

The TRS R consists of the following rules:

delete_3_in_aga₁(._2₂(X, Y)) -> delete_3_out_aga₂(X, Y)
delete_3_in_aga₁(._2₂(Y, Z)) -> if_delete_3_in_1_aga₁(delete_3_in_aga₁(Z))
if_delete_3_in_1_aga₁(delete_3_out_aga₂(X, W)) -> delete_3_out_aga₂(X, W)

The set Q consists of the following terms:

delete_3_in_aga₁(x0)
if_delete_3_in_1_aga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PERM_2_IN_GA₁, IF_PERM_2_IN_1_GA₁}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

PERM_2_IN_GA₁(._2₂(X, ._2₂(Y, []_0))) -> IF_PERM_2_IN_1_GA₁(delete_3_in_aga₁(._2₂(X, Y)))

The remaining Dependency Pairs were at least non-strictly be oriented.

IF_PERM_2_IN_1_GA₁(delete_3_out_aga₂(U, Z)) -> PERM_2_IN_GA₁(Z)

With the implicit AFS we had to orient the following set of usable rules non-strictly.

if_delete_3_in_1_aga₁(delete_3_out_aga₂(X, W)) -> delete_3_out_aga₂(X, W)
delete_3_in_aga₁(._2₂(Y, Z)) -> if_delete_3_in_1_aga₁(delete_3_in_aga₁(Z))
delete_3_in_aga₁(._2₂(X, Y)) -> delete_3_out_aga₂(X, Y)

Used ordering: POLO with Polynomial interpretation:

POL(._2₂(x₁, x₂)) = x₁ + x₂
POL(IF_PERM_2_IN_1_GA₁(x₁)) = x₁
POL(delete_3_in_aga₁(x₁)) = x₁
POL(if_delete_3_in_1_aga₁(x₁)) = x₁
POL(delete_3_out_aga₂(x₁, x₂)) = x₂
POL([]_0) = 1
POL(PERM_2_IN_GA₁(x₁)) = x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

IF_PERM_2_IN_1_GA₁(delete_3_out_aga₂(U, Z)) -> PERM_2_IN_GA₁(Z)

The TRS R consists of the following rules:

delete_3_in_aga₁(._2₂(X, Y)) -> delete_3_out_aga₂(X, Y)
delete_3_in_aga₁(._2₂(Y, Z)) -> if_delete_3_in_1_aga₁(delete_3_in_aga₁(Z))
if_delete_3_in_1_aga₁(delete_3_out_aga₂(X, W)) -> delete_3_out_aga₂(X, W)

The set Q consists of the following terms:

delete_3_in_aga₁(x0)
if_delete_3_in_1_aga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PERM_2_IN_GA₁, IF_PERM_2_IN_1_GA₁}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.